02 heat transfer

Integrated Training Program / Phase A – Heat Transfer Page 2 of 33

Copyright © 2004 International Human Resources Development Corporation

HEAT TRANSFER

Introduction

Upon completion of this course, the student will be able to:

1. Identify and explain the three types of heat transfer and the laws associated with

the three modes of heat transfer

2. Define thermal conductivity and thermal resistance

3. Calculate the rate transfer by conduction both for plane and cylindrical systems

4. Explain the overall heat transfer coefficient

5. Calculate rates of heat transfer using the overall heat transfer coefficient for plane

and cylindrical systems

6. Describe the different types of heat exchangers commonly used in the industry

7. Describe the operation of heat exchangers and evaluate related variables

associated with heat exchanger operation (LMTD, CMTD, heat transfer

coefficients, and sizing)

8. Calculate heat transfer rates in heat exchangers.



UNIT 1

HEAT TRANSFER

Introduction

Heat transfer and the different laws governing heat transmission are very important in

the design and operation of equipment like steam generators, furnaces, pre-heaters, heat

exchangers, coolers, evaporators and condensers in different industries.

The primary objectives of any heat transfer operation are:

1. To obtain the maximum heat transfer rate per unit surface compatible with

economic factors

2. Recovery of heat as in heat exchangers, recuperators, and regenerators

3. Minimizing heat losses by use of insulation

In a given operation, all three objectives may be important.

LESSON 1

Modes of heat transfer

1.1 Conduction

Conduction is the transfer of heat from one part of a substance to another under the

influence of a temperature gradient, without appreciable displacement of the molecules

of that substance. It is the transfer of energy from one molecule to an adjacent

molecule. Conduction is the only mechanism of heat flow in an opaque solid.

With transparent solids (e.g. glass), some energy is transmitted by radiation as well as

by conduction.

With gases and liquids, conduction may be supplemented by convection and radiation.

With flowing fluids, heat is transferred by conduction at right angles to the direction of

flow.

Examples of conduction are: heat transfer through walls, pipes, freezing of the ground

in winter, etc.

1.2 Convection

Convection involves the transfer of heat by the molecules of a substance moving from

one place to another and by mixing with each other. This mode of heat transfer

involves primarily fluids (liquids and gases).



The motion of the fluid may be due to density difference resulting from temperature

differences (natural convection), or the motion may be produced by mechanical

means, as in forced convection.

During heat transfer by convection, energy is also transferred simultaneously by

molecular conduction, and, in transparent media, by radiation.

Examples of heat transfer by convection are: cooling or heating of fluids in heat

exchangers, cooling of a cup of coffee by blowing over the surface, baking a cake in a

gas oven, etc.

1.3 Radiation

In contrast to the mechanisms of conduction and convection, where energy transfer

through a material medium is involved, heat may also be transferred through a vacuum.

This mode of heat transfer is electromagnetic radiation. A hot body emits radiant

energy in all directions. When this energy strikes another body, part of it may be

reflected, part may be transmitted through the body, and the remainder is absorbed and

quantitatively transformed into heat.

Examples of radiation are: transfer of heat from the sun to the earth, heating of tubes in

a furnace, etc.



LESSON 2

Laws of Heat Transfer

2.1 Conduction Heat Transfer

When a temperature gradient exists in a body, there is an energy transfer from the high

temperature region to the low temperature region – this is heat transfer by conduction.

Heat transfer by conduction is governed by Fourier’s Law, which for one-dimensional

cases is given as:

q = -k A dT/dx (1)

In the above equation,

q = the heat transfer rate in watts

A = the area involved in the heat transfer

dT/dx = the temperature gradient in the direction of heat flow

k = thermal conductivity of the material (values of k for

different materials can be found in the data book).

The negative sign in the equation is due to the fact that heat flows downhill on a

temperature scale. K has the units of watts per meter per 0C where heat flow is

expressed in watts.

Equation (1) is known as the Fourier’s Law of Heat Conduction

2.2 Convection Heat Transfer

When a heated solid is placed in air, heat is carried away from the solid surface. Close

to the surface of the solid, heat is lost by conduction, but as the distance increases from

the surface, heat transfer is by a convection process.

To express the overall effect of convection, the following equation is used:

q = hA (Tw - T) (2)

In the above equation, Tw and T are the temperatures at the wall and in the free stream

respectively. A is the area of heat flow. The above equation is also known as Newton’s

law of cooling.

The quantity h is called the convective heat transfer coefficient. This is also sometimes

called the film conductance because of its relationship to the conduction process in the

thin stationary layer of fluid at the solid surface.

Units of h: watts per square meter per 0C, when heat flow (q) is in watts.



2.3 Radiation Heat Transfer

In the radiation mode, heat transfer takes place by electromagnetic radiation (when there

is a temperature difference).

A black body (called an ideal radiator) emits energy at a rate proportional to the fourth

power of the absolute temperature of the body. When two bodies exchange heat by

radiation, the net heat exchange is then proportional to the difference in T4

q = (T14 - T2

4) (3)

The symbol is called the Stefan Boltzmann constant and has a value of

5.669 x 10-8 w/(m2 K4).

Equation (3) is called the Stefan Boltzmann’s law of thermal radiation and is applicable

only to black bodies. For other surfaces, the emisivity factor (F) and the geometric

shape factor (Fg) has to be added to the equation:

q = F Fg (T14 - T2

4)

(4)



∆x

q

q

LESSON 3

Steady State Conduction

3.1 Conduction in a plane surface:

If conduction heat transfer takes place at steady conditions through a plane surface

(plane wall), equation 1 can be written as:

k A

q =- ------- (T2 – T1) (5)

x

This may be shown diagrammatically:

T1 T2( T1 T2)



If more than one material is present, as in a multi-layered wall, the heat flow equation

becomes:

T1 - T4

q = ------------------------------------------

xA/(kA A) + xB/(kB A) + xC/(kC A)

Electrical Analogy:

Equation (5) can be written as:

Thermal Potential Difference

Heat Flow Rate = ---------------------------------------

Thermal Resistance

The above is analogous to Ohm’s Law in Electric circuit theory. The heat transfer rate is

the flow; and the combination of thermal conductivity, thickness of the material, and the

area represents the resistance to this flow; the temperature is the potential, or driving

force for heat flow. The heat conduction can then be represented as an electric analogy:

T1 RA T2 RB T3 RC T4

O O O O

xA/(kA A) xB/(kB A) xC/(kC A)

where RA , RB , and RC represents the three resistances in series.

q q

A B C

∆XA ∆XB ∆XC

1 2 3 4



L

3.2 Radial Systems (cylinders)

When heat is flowing through the sides of a cylindrical body of circular cross section (e.g. a pipe) the direction of heat flow is at all points radial and perpendicular to the axis; the cross section of the path of heat transfer is proportional to the distance from the center of the cylinder. In these cases, it will be shown below that the logarithmic-mean area is the proper average value.

Inner Radius =ri

Outer Radius =ro

Length = L

Inner wall temperature = Ti ; Outer wall temperature = To

Fourier’s Law for heat conduction for radial systems can be written as:

qr = -k Ar dT/dr (6)

The area for heat flow in the radial path is given as: Ar = 2rL

Substituting the expression for Ar, equation (6) becomes

qr = -2rL k dT/dr (7)

Noting that at r = ri , T = Ti and that at r = ro , T = To , equation (7) can be integrated (with q constant):

2kL (Ti – To) q = ------------------------- (8)

ln(ro/ri)

Equation (8) is the expression for heat transfer through a cylinder. The thermal resistance (Rth) in this case is:

ln(ro/ri) Rth = ------------- (9)

2kL



For a multi-layered cylindrical wall (e.g. for a pipe with several layers of insulation),

the expression for heat transfer becomes:

2L (T1 – T4)

q = ---------------------------------------------- (10)

ln(r2/r1)/kA + ln(r3/r2)/kB + ln(r4/r3)/kC

The electrical analogy will be represented as:

T1 RA T2 RB T3 RC T4

O O O O

ln(r2/r1)/2kAL ln(r3/r2)/2kBL ln(r4/r3)/2kCL

Example Problems (Steady State Conduction)

Example 1

A composite three-layered wall is formed of a 0.5 cm thick aluminum plate, a 0.25 thick

layer of sheet asbestos, and a 2.0 cm thick layer of rock wool; the asbestos is the central

layer. The outer aluminum surface is at 500oC and the outer rock wool surface is at

50oC. Determine the heat flow per unit area.



The thermal conductivities of the three materials are:

kal = 268.08 W/(m oK) ; kasb = 0.166 W/(m oK); krw = 0.0548 W/(m oK)

Solution:

The heat flow per unit area can be written as:

T1 - T4

q/A = ----------------------------------

x1/k1 + x2/k2 + x3/k3 )

(500 - 50) oC

q/A = -----------------------------------------------------------

(0.5x 10-2)m (0.25x 10-2)m (2.0 x 10-2)m

---------------------- + --------------------- + ----------------------

268.08 W/(m oK) 0.1660 W/(m oK) 0.0548 W/(m oK)

= 1184.08 W/m2 (answer)

q

q

A1 ASb Rw

0.5 0.25 2.0

50°C

500°C

1 2 3 4



Example 2

A 8 cm outer diameter (o.d.) steel pipe is covered with a 5 cm layer of asbestos, that is

covered in turn with a 5.0 cm layer of glass wool. Determine:

(a) the steady state heat transfer rate per linear meter of pipe, and

(b) the interfacial temperature between the asbestos and the glass wool if the pipe

outer surface temperature is 250 0C and the glass wool outer temperature is 40 0C.Given: kasb = 0.2 W/(m oK) and kgw = 0.055 W/(m oK)

Solution:

Since this is an example of a radial system, the equation heat flux can be written as:

2L (T1 – T3)

q = -----------------------------------

ln(r2/r1)/kAsb + ln(r3/r2)/kgw

The heat transfer rate per linear foot is q/A. Thus the above equation can be written as:

2 (250 – 40)

q/L = -------------------------------------------

ln(5.5/4.0)/0.2 + ln(10.5/5.5)/0.055

1319

= ------------------------- = 98.84 W/m (answer, a)

1.592+ 11.757

Since the heat transfer rate per linear foot (98.84 W/m) is known, the equation can be

written for a single layer (the glass wool layer):

2 (T2 – T3) 2 (T2 – 40) 2 (T2 – 40)

q/L = ---------------- = -------------------- = ---------------------

ln(r3/r2)/kgw ln(10.5/5.5)/0.055 11.757

(q/L) 11.757

(T2 – 40) = ---------------- = 184.95

2

T2 = 225 0C (answer, b)

Note: We could also have chosen the asbestos layer to find T2 since q/L is the same.



∆X

LESSON 4

Overall Heat Transfer Coefficient

In industrial processes, heat transfer usually takes place by a combination of conduction

and convection mechanisms. In these circumstances, the overall heat transfer rate is to

be determined. The convective heat transfer coefficient (h) in equation 2 (Newton’s law

of cooling) is then replaced with U (the overall heat transfer coefficient). Thus equation

2 becomes:

q = U * A * (T)overall (11)

Where (T)overall is the overall temperature difference

A is the area through which heat flows, and

U is the overall heat transfer coefficient

The units of U are the same as the units of h i.e W/(m2 0C)

a) Plane Wall

Let us take a plane wall as shown in the diagram below, made up of a homogeneous

material with a thermal conductivity k. One side of the wall is exposed to a fluid A at

temperature TA, and the other side to fluid B at temperature TB (TA > TB). The

thickness of the wall is x. The wall temperatures on the two sides of the wall are T1

and T2 (T1 > T2).

Fluid A Fluid B

K

hB

TA

TB

T1 T2



If we apply Newton’s law of cooling at the two surfaces this gives the heat transferred

per unit area:

q/A = hA (TA - T1) = hB (T2 - TB)

The equality holds because the heat transferred from one side of the wall is equal to the

heat transferred to the other side. Rearranging the equation,

(TA - T1) (T2 - TB)

q = ----------- = ------------

1/(hA A) 1/(hB A)

The terms 1/(hA A) and 1/(hB A) are thermal resistances due to the convective boundary

layer at the two surfaces of the wall.Hence the electrical analogy to this may be

represented by three resistances in series (Figure below), the conductive resistance

being x/kA from the conductive heat transfer equation (Equation 5).

q

TA T1 T2 TB

O O O O

1/(h1 A) x/(k A) 1/(h2 A)

The heat transfer rate through each layer of resistance of a plane wall is:

(TA - T1) (T1 - T2) (T2 - TB)

q = ----------- = ------------ = ------------

1/(hA A) x/(k A) 1/(hB A)

The conductive heat flow is exactly equal to convective heat flow on either side of the

plane wall. On combining the equations to find the heat transfer per unit area:

T1 - T2

q/A = ------------------------------- (12)

1/hA + x / k + 1/hB

Thus the overall heat transfer coefficient (U) for a plane wall becomes:

1 W

U = ------------------------------- (13) having units of ------------

1/hA + x / k + 1/hB m2 * 0K



T1

Ti

ro

ri

To

T2

b) Radial Systems:

q

Ti T1 T2 To

O O O

Ri Ra Ro

Let us consider the cylindrical system (e.g. pipe) shown above: Ti and To are the

temperature of the fluids flowing on the inner and outer sides of the system and ri and ro

are the inner and outer radii; T1 and T2 are the surface temperatures as shown. Heat

flow through the system can be represented as:

(T)overall Ti - To

q = ----------------- = --------------------

Rthermal Ri + Ra + Ro

In the above equation,

Ri = inner convective thermal resistance = 1/(hiAi) = 1/ (2riLhi)

ln (r2/r1)

Ra = conductive thermal resistance = -------------

2ka L

Ro = outer convective thermal resistance = 1/(hoAo) = 1/ (2roLho)

L = the length of the cylindrical system (pipe)

Similar to the analysis of the plane wall, the heat transfer rate through each layer of

resistance of the cylindrical system can be shown as:

Ti - T1 T1 - T2 T2 - To

q = ----------- = -------------------------- = --------------

1/(hiAi) ln(r2/r1) /(2kaL) 1/(hoAo)



This is because the heat flow through each resistance layer is exactly the same. Thus

we can write the equation of the heat transfer through the system as:

Ti - To

q = -------------------------------------------------------------- (14)

1/ (2riLhi) + ln(r2/r1) /(2kaL) + 1/ (2roLho)

The overall heat transfer coefficient for a cylindrical system is based on either the inner

surface area (Ai) or the outer surface area (Ao) of the cylinder (or pipe). Therefore, we

can write :

Ti - To

q = Ui Ai (Ti - To) = Uo Ao (Ti - To) = ------------

R

1

Uo = -------------------------------- (15)

r2 r2 ln (r2/r1) 1

----- + ---------------- + ------

r1hi ka ho

1

Ui = -------------------------------- (16)

1 r1 ln (r2/r1) r1

----- + ---------------- + ------

hi ka r2 ho

Uo and Ui are the overall heat transfer coefficients based on the outer and inner surface

areas respectively.



Example problems (Overall Heat Transfer Coefficient)

Example 3

A rectangular steel tank is filled with a liquid at 700C and exposed along the outside to

air at 200C. The inner and outer heat transfer coefficients are hi = 23 W/(m2 K) and ho =

8 W/(m2 K). The tank wall is 1 cm. Thick mild steel with thermal conductivity k = 45

W/(m*K) and this is covered with a 3 cm. Layer of glass wool insulation ( k = 0.04

W/m*K). Determine: (a) the overall heat transfer coefficient and (b) the heat transfer

rate per square meter area.

Solution:

Write the expression for the overall heat transfer coefficient for plane surfaces with two

conductive layers:

1

U = -----------------------------------------

1/hi +x1/k1 + x1/k1 + 1/hi

1

or U = -----------------------------------------------------

1/23 + (1x10-2)/45 + (3x 10-2)/0.04 + 1/8

1

or U = ---------------------------------------------

+ 0.00022 + 0.75 + 0.125

= 1/0.9187

= 1.08 W/(m2 K) (answer a)

(Note that the thermal resistance of the steel wall is negligible and that of the glass wool

is predominant)

The heat transfer rate per square meter:

Q/A = U * T = (1.08)(70 - 20)

= 54 W/ m2 (answer b)



Example 4

Water at 980C flows through a 2 inch schedule 40 horizontal steel pipe (k = 54 W/(m 0C) and is exposed to atmospheric air at 200c.Calculate the overall heat transfer

coefficient for this situation, based on the outer area of the pipe. The inner and outer

convective heat transfer coefficients are: hi = 1961 W/(m2 0C) and ho = 7.91 W/(m2 0C).

Given: the I.D. and O.D. of the pipe are 0.0525 m and 0.06033m respectively.

Solution:

Write the expression for the overall heat transfer coefficient for a radial system based on

the outer surface area:

1

Uo = -----------------------------------

r2 r2 ln (r2/r1) 1

----- + ---------------- + ------

r1hi ka ho

1

Uo = -----------------------------------------------------------------------

0.06033 0.06033 ln (0.06033/0.0525) 1

---------------- + ------------------------------------ + ------------

(0.0525)(1961) 54 7.91

1 1

Uo = ------------------------------------------- = ----------------

0.00059 + 0.00021 + 0.1264 0.1272

Uo = 7.86 W/ (m2 * 0C) (answer)

(note: the overall heat transfer coefficient is controlled by ho ).



UNIT 2

HEAT EXCHANGERS

There are many different types of industrial heat exchange equipment (heat exchangers).

However, the one characteristic common to most heat exchangers is the transfer of heat

from a hot fluid to a cold fluid with the two fluids being separated by a solid boundary.

The proper operation and maintenance of heat exchangers is crucial to the efficient

operation of any process plant.

LESSON 1

Types of Heat Exchangers:

1. Double Pipe Heat Exchanger

This is the simplest type of heat exchanger and is essentially two concentric pipes with

one fluid flowing through the center pipe while the other fluid flows co-currently or

counter-currently in the annular space.The following figure shows a diagram of a

double pipe exchanger with counter-flow arrangement.

2. Shell and Tube Heat Exchanger

When the heat required transfer surface is large, the recommended type of heat

exchanger is the Shell and Tube variety. In this type of cooler or heater, large heat

transfer surface can be achieved economically and practically by placing tubes in a

bundle; the ends of the tubes are mounted in a tube sheet. The tube bundle is then

enclosed in a cylindrical casing (the shell) through which one fluid flows around and

through the tube bundle.

Fluid flowing through the tubes enters a header or channel where it is distributed

through the tubes in parallel flow and leaves the unit through another header. Either the

hot or the cold fluid may flow in the shell of the exchanger surrounding the tubes.

Parallel flow through all tubes at a low velocity produces low heat transfer rates. For

higher rates of heat transfer, multi-pass operation may be used. In multi-pass operation,

the fluid in the tubes is diverted by baffles within the distribution header. Higher

velocity and thus higher heat transfer rates are thus achieved.



Baffles on the shell side divert the flow of the shell fluid into a path predominantly

across the tubes of the exchanger. This increases turbulence and enhances heat transfer

rates. Baffles in the shell are mostly semi-circular discs of sheet metal accommodating

the tubes.

Flow in a double pipe heat exchanger (counter current)



Cold fluid in hot fluid out

Hot fluid in cold fluid out

(a)

Cold fluid in hot fluid in

Cold fluid out hot fluid out

(b)

Shell and tube heat exchangers: (a) 1 shell pass and 1 tube pass (1-1 exchanger). (b)

1 shell pass and 2 tube passes (1-2 exchanger)



t2

t2

t1

T2

T1

T1

T2

T2 ∆T2

∆T1

t2

t1

t1

t2

LESSON 2

Types of Flow in Heat Exchangers

2.1 Co-current Flow

Both fluids flow in the same direction

T1

T1

T2

T2 T1

t1

Length of exchanger

2.2 Counter-current Flow

One fluid flows in the opposite direction of the other

Length of exchanger

t1



LESSON 3

Determination of Heat Transfer for Heat Exchangers

The rate of heat transfer for heat exchangers is given by the following three

equations:

i) Heat lost by the hot fluid:

q = mh * Cp * ∆Th (17)

ii) Heat gained by the cold fluid:

q = = mc * Cp * ∆Tc (18)

iii) Heat transferred between the fluids:

q = U * A * ∆TLM (19)

The q for each equation is equal as the heat lost by the hot fluid is equal to the heat

gained by the cold fluid and is also equal to the heat transferred between the two fluids.

In the above equations,

mh and mc are the mass flow rate of the hot and cold fluids

Cp is the specific heat

∆Th = the temperature difference of the hot fluid as it enters and leaves the exchanger

∆Tc = the temperature difference of the cold fluid as it enters and leaves the exchanger

U = the overall heat transfer coefficient, and

A = the surface area of the exchanger

∆TLM = the Log Mean Temperature Difference (LMTD) or

the logarithmic average difference between the two streams and given by:

( ∆T2 - ∆T1)

∆TLM = ----------------------- (20)

ln (∆T2 /∆T1 )

where ∆T2 and ∆T1 are the temperature differences between the two fluids at each end

of the exchanger. These temperature differences are also called the terminal temperature

differences.



Usually ∆T2 is the higher of the two and is also referred to as GTTD (Greatest Terminal

Temperature Difference). ∆T1 is then referred to as LTTD (Least Terminal

Temperature Difference).

It is important to have a consistent basis for the overall heat transfer coefficient and the

area. In most cases the outside surface area is used . Thus :

q = Uo * Ao * ∆TLM (21)

The above equation applies for exchangers with pure counter current flow. However, in

most commercial heat exchangers of the shell and tube type (with more than one tube

pass), the flow of fluids is neither co-current nor counter-current. The flow is really a

complex pattern of mixed flows.

Thus in a shell and tube heat exchanger, ∆TLM is used along with a correction factor (F)

to account for the geometry of the exchanger and for the fact that flow is not truly co-

current or counter-current. Thus the equation for heat transfer for a shell and tube heat

exchanger can be written as:

q = Uo * Ao * F * ∆TLM (22)

where F is correction factor. The product of F and ∆TLM is termed the Corrected Mean

Temperature Difference (CMTD). So the equation for heat transfer becomes:

q = Uo * Ao * CMTD

where CMTD = F * LMTD and LMTD = ∆TLM

The correction factors are determined from the LMTD Correction Factor tables found in

the GPSA Handbook (page 9-4, 9-5). Most heat exchangers in service require a

correction factor as most of them have more than one tube pass.

The correction factors are given in the handbook in terms of two dimensionless ratios:

(a) The temperature ratio: R = (T1 – T2) / (t2 – t1) (23)

(b) The effectiveness ratio: P = (t2 – t1) / (T1 – t1) (24)

In both the above expressions the higher case (T) is used for the shell fluid temperature

and the lower case (t) is used for the tube side fluid.

In the GPSA hand book, Section 9 deals with industrial heat exchangers. Typical

values of the overall heat transfer coefficient for heat exchanger use in the natural gas

industry are given in Figure 9-9.

All heat exchangers in use have a specification sheet with all the relevant operational

data. An example of a heat exchanger specification data sheet is given in Figure 9-12 in

the hand book.



Sample Problems on Heat Exchangers:

Example 1

Water at a flow rate of 550 kg/hr is cooled in a double pipe heat exchanger from 50oC

to 25oC by a cool brine solution which enters at 5oC and exits at 12oC. If the overall heat

transfer coefficient is 1020 W/(m2 oC), what is the heat exchanger area required for (a)

co-current flow and (b) counter-current flow ?

(Given Cp for water is 4176 J / (kg oC).

Solution:

The heat transfer rate (q) can be found from the data for the water stream:

q = m*Cp * ∆Th

= 550 kg/hr *4176 J/(kg oC) * (50 - 25) oC * 1/3600 sec./ hr

= 15950 W (Watts)

The heat transfer rate can also be written as q = U * A * ∆TLM

Solving for the heat transfer area A,

q

A = ---------------

U * ∆TLM

( ∆T2 - ∆T1)

Now ∆TLM = -----------------------

ln (∆T2 /∆T1 )

∆T2 = GTTD = 50 - 5 = 45 oC and ∆T1 = LTTD = 25 - 12 = 13 oC

Thus

( 45 - 13)

∆TLM = ------------------- = 25.77 oC

ln (45 /13 )

15950 W

So A = ---------------------------------- = 0.607 m2 (answer a)

1020 W /(m2 oC) * 25.77 oC



(b) For counter-current flow conditions,

∆T2 = GTTD = 50 - 12 = 38 oC and ∆T1 = LTTD = 25 - 5 = 20 oC

and

( 38 - 20)

∆TLM = ------------------- = 28.04 oC

ln (38 /20)

15950 W

So A = --------------------------------- = 0.558 m2 (answer b)

1020 W /(m2 oC) * 28.04 oC

(note: for the same heat duty, the counter current arrangement requires less area).



Example 2

Water at a rate of 68 kg/min is heated in a counter current double pipe heat exchanger

from 35 oC to 75 oC by a hot oil stream entering at 110 oC and leaving at 75 oC. The

heat capacity of the water is 4.19 kJ/kg oC and the heat capacity of the oil is 1.9 kJ/kg oC. The overall heat transfer coefficient for the exchanger is 320 W/m2 oC. What is the

required flow of the oil and the required surface area of heat transfer?

Solution:

The heat transfer rate is determined from the heat absorbed by the water:

q = mw * Cp(w) * Tw

= 68 kg/min * 1/(60 sec/min) * 4.19 kJ/kg oC * (75 – 35) oC

= 189.95 kW = 189950 W = 189 .95 kJ/sec = 189950 J/sec.

This amount of heat is equal to the heat given up by the oil. Therefore,

q = 189.95 kW = moil * Cp(oil) * Toil

189.95 kW 189.95 kW

moil = ------------------ = -----------------------------

Cp(oil) * Toil 1.9 kJ/kg oC * (110 –75) oC

= 2.86 kg/sec = 171.6 kg/min. (answer)

Calculating heat transfer area:

LMTD for the exchanger is given by:

(75 – 35) - (110 – 75)

TLM = ----------------------------------- = 37.44 oC

ln (( 75 – 35) /(110 – 75))

q 189950 J/sec

Area of the exchanger, A = ----------- = -------------------------

U * ∆TLM 320 W/m2 oC * 37.44oC

Therefore A = 15.85 m2 (answer)



Example 3

Instead of the double pipe heat exchanger of the previous example, it is desired to use a

shell and tube exchanger with water making one shell pass and the oil making two tube

passes. Calculate the area required for this exchanger, assuming that the overall heat

transfer coefficient remains at 320 W/m2 oC.

Solution:

To solve this problem, we have to determine a correction factor for the LMTD

calculated on the basis of a counter current exchanger. As already mentioned the

correction factor can be found from pages 9-4 and 9-5 in the GPSA Hand Book.

For this problem,

R = (T1 – T2) / (t2 – t1) = (35 – 75) / (75 – 110) = 1.143

P = (t2 – t1) / (T1 – t1) = (75 – 110) / (35 – 110) = 0.467

The correction factor (F) for a 1 shell pass and 2 tube passes exchanger (1-2 exchanger)

is found to be: F = 0.81

Thus writing the expression for the area of the exchanger to include the correction

factor:

q 189.95 kJ/sec

A = ----------- = ------------------------------------------

U *F *∆TLM 320 W/m2 oC * 0.81 *37.44oC

A = 19.53 m2 (answer)



Example 4

A shell and tube heat exchanger operates with 2 shell passes and 4 tube passes. The

shell fluid is ethylene glycol (specific heat = 2.5 kJ/kg oC) which enters at 140 oC and

leaves at 80 oC with a flow rate of 4500 kg/hr. Water flows in the tubes, entering at 35 oC and leaves at 85 oC. The overall heat transfer coefficient for this arrangement is 850

W/m2 oC. Calculate the flow rate of water required and the area of the exchanger.

Solution:

To calculate the heat transfer rate, we take the glycol stream as all data for this stream is

known:

q = mg * Cp(g) * Tg

= (4500 kg/hr * 1/3600 s/hr ) * 2.5 kJ/kg oC * (140 – 80)oC

= 187.5 kW = 187.5 kJ/sec = 187500 J/sec

Find the flow rate of the water:

187.5 kW 187.5 kW

mw = ------------------ = -----------------------------

Cp(w) * Tw 4.18 kJ/kg oC * (85 –35) oC

= 0.89713 kg/sec. = 3229.7 kg/hr.

Find the area of the exchanger:

Since this is a shell and tube exchanger, we need correction factor for the LMTD

(∆TLM).

(140 – 85) – (80 – 35)

∆TLM = -----------------------------------

ln ((140 – 85) – (80 –35))

= 49.85 oC

Find the correction factor for this exchanger:

R = (T1 – T2) / (t2 – t1) = (140 – 80) / (85 – 35) = 1.2

P = (t2 – t1) / (T1 – t1) = (85 – 35) / (140 – 35) = 0.476



Correction factor F from hand book : F = 0.95

If we write the expression for the area of the 2-4 exchanger,

q 187500 J/sec

A = ----------- = -------------------------

U *F *∆TLM 850 W/m2 oC * 0.95 *49.85oC

A = 4.66 m2 (answer)



Lesson 4

Fixed and Floating Head Heat Exchangers

Heat exchangers with fixed tube sheets are difficult to clean, both on the shell side as

well as in the tube side. Also there is virtually no provision for differential thermal

expansion of the tube bundle and the shell which can be caused by large temperature

differentials between the shell and tube fluids.

To allow provisions for easy removal of the tube bundle for cleaning and to allow for

thermal expansion a floating head heat exchanger is used (Fig., page 25 ). In this type

of heat exchanger, one of the tube sheets is independent of the shell so that the entire

tube bundle can be removed for cleaning of the shell and the outside of the tubes. This

also allows for differential expansion between the tube s and the shell.

LESSON 5

routing of the fluids in a shell and tube heat exchanger

If one of the fluids fouls up the surfaces much more rapidly than the other, it should be

routed through the tubes, since the inside of the tubes can be cleaned without removing

the tube bundle from the shell.

If both the fluids are equally non-fouling, and one is under high pressure, it should flow

inside the tubes to avoid the expense of a high pressure shell.

When only one of the fluids is corrosive, it should flow inside the tubes to avoid the

expense of special metal for both shell and tubes.

If one of the fluids is much more viscous than the other, it may be routed through the

shell to increase the overall heat transfer coefficient. At times, because of the limitation

in available pressure drop, it is necessary to route a given stream through the shell.



A Floating Head 2 – 4 Shell and Tube Heat Exchanger

Extended Surface Tubing



LESSON 6

Extended Surface Heat Exchangers

If heat exchange occurs between two fluids where one fluid has a very high resistance to

heat transfer in comparison to the other, the higher resistance fluid “controls” the rate of

heat transfer. Such cases occur, for example, in the heating of air by steam or in heating

of a very viscous oil, flowing in laminar flow, by a molten salt mix.

The relative magnitude of the heat transfer coefficient is about 60 W/m2 oC for the oil or

air, compared with 12000 W/m2 oC for the steam or salt (200 times greater).

Consequently, the overall coefficient is essentially equal to the individual coefficient for

the air or oil (close to 60 W/m2 oC); the capacity of a unit area of heating surface will be

low, and many lengths of tube will be required to provide a reasonable capacity.

In order to compensate for the high resistance of the oil or air, the heat transfer surface

exposed to these fluids may be increased by extension of the surface, as in the addition

of fins to the outside of the tubes . The fins are referred to as extended surfaces. These

extended surfaces increase the heat transfer area substantially in a given amount of

space. A common example of extended surface heat exchanger is the automobile

radiator.

02 heat transfer

Documents

transfer of heat

heat transfer operation

heat transfer page

modes of heat transfer

heat transfer rates

rates of heat transfer

types of heat transfer

examples of heat transfer